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I have been thinking the relation between two stable groups formed as follow:

1) $GL(2,\mathbb R)\subseteq GL(3,\mathbb R)\subseteq\cdots GL(n,\mathbb R)$, $G(\infty,\mathbb R)=\cup_{n=2}^{\infty} GL(n,\mathbb R)$

2)$GL(3,\mathbb R)\subseteq GL(5,\mathbb R)\subseteq\cdots GL(2n+1,\mathbb R)$, $H(\infty,\mathbb R)=\cup_{n=1}^{\infty}GL(2n+1,\mathbb R)$

It is clear that $H(\infty,\mathbb R)$ is a subgroup of $G(\infty,\mathbb R)$. Do you think that they might be isomorphic as a groups? If they are not, then what is the proper approach to explain them?

I can write a map from $G(\infty,\mathbb R)$ to $H(\infty,\mathbb R)$, for instance if $A\in GL(2,\mathbb R)$ and I want to send it to $GL(5,\mathbb R)$, then the map take $A$ and send it to $diag(A,I_{3})$. This map injective homomorphism, but it is nor surjective!

I do not know maybe they are not isomorphic, but in this case how can convince myself?

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  • $\begingroup$ It sounds trivial. If we have an inductive limit of $(G_n)$ with arrows $f_n:G_n\to G_{n+1}$, then it's canonically isomorphic to the inductive limit of $(G_{2n+1})$ with the maps $f_{2n+2}\circ f_{2n+1}$. It works in any category. $\endgroup$
    – YCor
    Commented May 6, 2017 at 20:24

1 Answer 1

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Define the map $i\colon H(\infty) \to G(\infty)$ by the sequence $i_k$ in the following commutative diagram (each vertical arrow is the obvious isomorphism): $$\require{AMScd}\require{color}\require{xcolor} \begin{CD} GL(2) @>>> GL(3) @>>> GL(4) @>>> GL(5) \cdots \\ @. @A{i_3}AA @. @A{i_5}AA \\ @. GL(3) @>>> {\color[gray]{.7}GL(4)} @>>> GL(5) \cdots \end{CD} $$

Define the map $p\colon G(\infty) \to H(\infty)$, by the sequence $p_k$ in the following commutative diagram, where the $p_{2k+1}$ are the obvious isomorphisms and (due to the limitations of AMScd diagram drawings) $p_{2k}$ should be interpreted as the isomorphism on to the image of the obvious inclusion ${\color[gray]{.7}GL(2k)} \to GL(2k+1)$: $$\require{AMScd}\require{color}\require{xcolor} \begin{CD} GL(2) @>>> GL(3) @>>> GL(4) @>>> GL(5) \cdots \\ @VV{p_2}V @VV{p_3}V @VV{p_4}V @VV{p_5}V \\ {\color[gray]{.7}GL(2)} @>>> GL(3) @>>> {\color[gray]{.7}GL(4)} @>>> GL(5) \cdots \end{CD} $$

The maps $i$ and $p$ are mutually inverse group homomorphisms (e.g., check for each element separately), thus each is a bijective homomorphism, meaning that $H(\infty) \cong G(\infty)$.

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  • $\begingroup$ Thank you so much for your prompt answer. I was thinking the diagrams for a while and I definitely agree with you that i_{3} and i_{5} are isomorphisms of groups and similarly p_{3} and p_{5} too, but if the inclusion mao GL(2) to GL(3) is i_{2,3}, then why i_{2,3} compose p_{2} is surgective? Thank you in advance. $\endgroup$
    – user101496
    Commented Sep 8, 2016 at 13:50
  • $\begingroup$ @user101496, there is no requirement that each of the maps in the $p_{2k+1}$, $i_{2k+1,2k}\circ p_{2k}$ sequence to be surjective for the corresponding map $p\colon G(\infty) \to H(\infty)$ to be surjective. Just pick any element $h\in H(\infty)$ and check that there exists a pre-image $g\in G(\infty)$ such that $p(g) = h$. $\endgroup$
    – Igor Khavkine
    Commented Sep 8, 2016 at 14:54
  • $\begingroup$ Thank you again Igor for your explanation. I agree with you so there is no problem with the surjectivity in this case. For the injection: I think we need to define the given maps on equivalence classes otherwise we might get more than one preimage of some elements in G(\infty). For instance the preimage of drag(A,1)\in GL(3) is A and drag(A,1), but if we think that A and drag(A,1) in the same class, then there is no problem anymore. If I am thinking wrong please let me know. Thank you again for your time to explain my questions with diagrams. That might take some time to type them. $\endgroup$
    – user101496
    Commented Sep 8, 2016 at 16:11
  • $\begingroup$ note that drag(A,1) should be diag(A,1) $\endgroup$
    – user101496
    Commented Sep 8, 2016 at 16:18
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    $\begingroup$ @user101496, so you have already have a solution for checking injectivity. I'm not sure that whether you are still missing anything (other than perhaps confidence with the subject). To build up more intuition/confidence I suggest that you look at more examples of inductive limits, of which this is just a special case. $\endgroup$
    – Igor Khavkine
    Commented Sep 8, 2016 at 21:33

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